physics 08 atom

Nuclei

 1 .(a) Two stable isotopes of lithium  63Li and 73Li have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.

(b) Boron has two stable isotopes   105B and  115B Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of .  105B and  115B.

(a) Mass of  63Li   lithium isotope , m1 =6.01512 u

Mass of   73Li  lithium isotope , m   =    7.01600 u

Abundance of  63Li  ,  n1 =7.5 % 

Abundance of  73Li  ,  n = 92.5%

The atomic mass of lithium atom is given as:

m = m1n1 +m2n2 /  n1+n2

= 6.01512 ×7.5 +7.01600 ×92.5  /  92.5 + 7.5

= 6.940934 u.

(b) Mass of   105B boron isotope  105B , m = 10.01294 u

Mass of boron isotope  115B , m = 11.00931 u

Abundance of  105B , n1  = x  %

Abundance of  115B , n=(100 -x) %

Atomic mass of boron, m = 10.811 u

The atomic mass of boron atom is given as: m =m1n1 +m2n2 / n1+n2

10.811 =10.01294 ×x +11.00931 ×( 100 – x)  /  x +100 -x

1081.11 = 10.01294 x  +  1100.931 – 11.00931 x

∴x = 19.821 /0.99637  =19.89 %

And 100 -x =  80.11%

Hence, the abundance of is 105B 19.89% and that of  115B  is 80.11%

 2:The three stable isotopes of neon:  2010Ne,  2110Ne   and ,2210Ne,have respective abundances of  90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.

Solution 2:
Atomic mass of  2010Ne , m=19.99u

Abundance of 2010Ne , η1  = 90.51%

Atomic mass of 2110Ne , m =20.99u

Abundance of 2110Ne, η=0.27%

Atomic mass of 2210Ne, m3=21.99u

Abundance of  2210Ne,η3 =9.22%

The average atomic mass of neon is given as:

m= m1n1 +m2n2+m3n3 / η+ η+  η3

=19.99 ×90.51 + 20.99 ×0.27 +21.99 × 9.22 / 90.51+ 0.27 +9.22

= 20.1771 u

3.Obtain the binding energy (in MeV) of a nitrogen nucleus ( 147N ) , given m( 147N) =14.00307u

Atomic mass of  ( 7N14 ) nitrogen, m =14.00307u

A nucleus of 7N14 nitrogen contains 7 protons and 7 neutrons. 

Hence, the mass defect of this nucleus, Δm=7mH +7mn – m

Where,

Mass of a proton, mH   = 1.007825  u

Mass of a neutron,mn  = 1.008665 u

Δm = 7 × 1.007825 + 7×1.008665 – 14.00307

= 7.054775 + 7.06055 − 14.00307

= 0.11236 u

But  1 u = 931.5 MeV /c2

Δm = 0.11236× 931.5 MeV /c2

Hence, the binding energy of the nucleus is given as:

Eb = Δmc2

Where, c = Speed of light 

E = 0.11236 × 931.5  ( MeV /c) × c2

=104.66334MeV

Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.

4.Obtain the binding energy of the nuclei  5626Fe and  20983Bi and in units of MeV from the following data:

m(5626Fe ) = 55.934939 u    m( 20983Bi  ) =208.980388 u 

Atomic mass of  (5626Fe )  , m=55.934939 u

(5626Fe )  nucleus has 26 protons and ( 56-26) =30 neutrons

Hence, the mass defect of the nucleus, Δm =26 ×mH+ 30 ×mn – m1

Where,

Mass of a proton, mH= 1.007825 u

Mass of a neutron, m= 1.008665 u

Δm =26 × 1.007825 + 30 × 1.008665  -55.934939

= 26.20345 + 30.25995 − 55.934939

= 0.528461 u

But 1 u = 931.5 MeV /c2

Δm = 0.528461 × 931.5 MeV /c2

The binding energy of this nucleus is given as:

Eb1 =Δmc2

Where, c =  Speed of light

∴ Eb1  = 0.528461 931.5 ( MeV /C2) × C2  

=492.26MeV

Average binding energy per nucleon = 492.26/56 =8.79 MeV

Atomic mass of    20983Bi, m=208.980388 u

 20983Bi  nucleus has 83 protons and (209 − 83) 126 neutrons.

Hence, the mass defect of this nucleus is given as:

Δm’ = 83 ×mH +126 ×mn -m2

Where,
Mass of a proton, mH = 1.007825 u

Mass of a neutron, m= 1.008665  u

Δm’ =83 ×1.007825  + 126 × 1.008665 – 208.980388 

=83.649475 + 127.091790 − 208.980388

= 1.760877 u

But 1 u = 931.5  MeV /c2

Δm’ = 1.760877 ×931.5  MeV /c2

Hence, the binding energy of this nucleus is given as:

Eb2 =Δm’ c’

= 1.760877 × 931.5  ( MeV /c) ×c2

=1640.26MeV

Average binding energy per nucleon = 1640.26/209  = 7.848 MeV

5. A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of 2963Cu atoms (of mass 62.92960 u).

Mass of a copper coin, m’  =3g 

Atomic mass of  29 Cu 63  atom, m = 62. 92960 u

The total number of 29 Cu 63  atoms in the coin, N = NA ×m’/ Mass number

Where,
NA = Avogadro’s number = 6.023 × 1023 atoms /g

Mass number = 63 g

∴N=6.023 × 1023  × 3/63 = 2.868 ×10 22 

29 Cu 63   nucleus has 29 protons and (63-29) 34 neutrons

∴ Mass defect of this nucleus, Δm’ = 29 × mH + 34× mn -m

Where,
Mass of a proton,mH= 1.007825 u

Mass of a neutron,mn =1.008665 u

∴ Δm’ = 29 × 1.007825 +34 × 1.008665 – 62.9296 

=0.591935 u

Mass defect of all the atoms present in the coin, ∆ m = 0.591935 × 2.868 × 1022

= 1.69766958 ×10 22 u

But 1 u =  931.5  MeV /C2

∴ Δm = 1.69766958 × 1022  × 931.5  MeV /C2

Hence, the binding energy of the nuclei of the coin is given as:

Eb =Δmc2

= 1.69766958  × 10 22 ×931.5 (  MeV /c2) ×c2

= 1.581  ×1025 MeV

But 1 MeV = 1.6 × 10 -13J

E= 1.581 × 1025  × 1.6 ×10 -13

= 2.5296  ×10 12J

This much energy is required to separate all the neutrons and protons from the given coin,

6.Write nuclear reaction equations for

(i) α – decay of 226 /88  Ra  

(ii) α – decay of  242/94 Pu

iii)β – decay of  32/15p

(iv) β – decay of  210/83 Bi

Solution 6:
α is a nucleus of helium( 2He 4 ) and β is an electron (e–  for β–  and e+ for  β+)  In every α  decay, there is a loss of 2 protons and 2 neutrons. In every β+ decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every β  decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus. For the given cases, the various nuclear reactions can be written as:

(i)  88Ra 226  →  86Rn222 2He 4

(ii)94Pu 242 92U238 + 2He 4

                                         _

(iii)15P 32 16S32 + e + v

                                              _

(iv)83B 210 84PO210 + e++ v

(V) 6C 115B11 + e++ v

(vi) 43Tc9742MO97 + e++ v

(vii) 54Xe120 + e+  →  53I120  + v

7.  A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to  a) 3.125%,    b) 1%    of its original value?

Half-life of the radioactive isotope = T years
Original amount of the radioactive isotope = N0

(a) After decay, the amount of the radioactive isotope = N
      It is given that only 3.125% of N0 remains after decay. Hence, we can  write:

N/N0 = 3.125% = 3.125/100 =1/32

but N/N0 = e -λl

Where , λ = Decay constant   t = time

∴ – λt =1/32

– λt = ln l  – ln 32

– λt = 0 -3.4657

t= 3.4657/λ

since λ = 0.693/T

∴ t= 3.466/(0.693/T)    ≈ 5T years

Hence, the isotope will take about 5T years to reduce to 3.125% of its original value

(b) After decay, the amount of the radioactive isotope = N
It is given that only 1% of N0 remains after decay. Hence, we can write:

N/N0  = 1% =1/100

but N/N0  = e -λlt

∴e -λlt  =1/100

-λ t = ln 1  – ln 100

-λ t =0 -4.6052

t= 4.6052 /λ

Since, λ = 0.639/T 

∴t= 4.6052 / ( 0.693/T) = 6.645 T year

Hence, the isotope will take about 6.645T years to reduce to 1% of its original value.

8.The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive  6C 14present with the stable carbon isotope 6C12 . When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity  begins to drop. From the known half-life (5730 years) of 6 C14 , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of 6C14 dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.

Decay rate of living carbon-containing matter, R = 15 decay/min
Let N be the number of radioactive atoms present in a normal carbon- containing matter.
Half life of  6C14 .T1/2 =5730 years
The decay rate of the specimen obtained from the Mohenjodaro site:
R’ = 9 decays/min
Let N’ be the number of radioactive atoms present in the specimen during the Mohenjodaro period.
Therefore, we can relate the decay constant,λ and time, t as:

N’/N =R’/R =e-λt

e-λt =9/15 =3/5

-λt = log e 3/5 = -0.5108

∴t = 0.5108/λ

but λ =0.693/T1/2  =0.693 /5730

∴t=0.5108/(0.693 /5730) = 4223.5years

Hence, the approximate age of the Indus-Valley civilization is 4223.5 years

9. Obtain the amount of 27Co60 necessary to provide a radioactive source of 8.0 mCi strength. The half-life of  27Co60  is 5.3 years.

The strength of the radioactive source is given as:

dN/dt = 8.0 m Ci

=8 × 10 -3  ×3.7 ×10 10 

= 29.6 × 10 decay/s

Where,
N = Required number of atoms

Half-life of  27Co60  , T 1/2   =5.3 years

= 5.3 × 365 ×24 ×60 ×60

=1.67 ×108 s

For decay constant λ , we have the rate of decay as:

dN/dt = λN 

Where

λ= 0.693 /T1/2 =0.693  / 1.67 × 10-1

∴N= ( 1/λ)  ( dN/dt) 

=29.6 × 10 7  / (  0.693/1.67 × 108)= 7.133 ×10 16 atoms

for 27Co60 

Mass of 6.023 × 1023 (Avogadro’s number) atoms = 60 g

Mass of  7.133 10 16 atoms = 60 ×7.133 × 10 16  / 6.023 ×10 23  =7.106 × 10 -6 g

Hence, the amount of 27Co60 necessary for the purpose is 7.106 × 10 -6 g

10 .The half-life of  38 Sr 90  is 28 years. What is the disintegration rate of 15 mg of this isotope?

Half life of  38 Sr 90 , t 1/2  =28years

=28 ×365 ×24 ×60 ×60 

=8.83 ×10 s

Mass of the isotope, m = 15 mg
90 g of 90 Sr38  atom contains 6.023 × 1023 (Avogadro’s number) atoms.

Therefore, 15 mg of 90 Sr38  contains:

6.023 × 10 23 × 15 × 10 -3   /90   i,e ., 1.0038 ×10 20 Number of atomms

Rate of disintegration, dN/dt = λN

Where,  

λ = decay constant =0.693 / 8.83 ×10  s-1

∴dN/dt = 0.693 ×1.0038 ×10 20 /8.83 ×10 8  =7.878 ×10 10 atoms/s

Hence, the disintegration rate of 15 mg of the given isotope is 7.878 ×10 10 atoms/s

11.Obtain approximately the ratio of the nuclear radii of the gold isotope  79 Au197 and the silver isotope   47 Ag 107.

Solution 11:
Nuclear radius of the gold isotope  79Au197  =R Au

Nuclear radius of the silver isotope 47Ag107 = R Ag

Mass number of gold AAu = 197 

Mass number of silver,AAg  =107 

The ratio of the radii of the two nuclei is related with their mass numbers as:

(RAu / RAg ) = (AAu / AAg) 1/3 = 1.2256

Hence, the ratio of the nuclear radii of the gold and silver isotopes is about 1.23.

12. Find the Q-value and the kinetic energy of the emitted α-particle in the α decay of (a) 88 Ra 226 and (b)  86 Rn 220

Given  m (  88Ra 226 ) = 226.02540u , m (222 Rn89 )= 222.01750u,

m(86Rn220  ) = 220.01137u,m( 84po216 )= 216.00189u

(a) Alpha particle decay of 88Ra 26 emits a helium nucleus. As a result, its mass number reduces to (226 − 4) 222 and its atomic number reduces to (88 − 2) 86. This is shown in the  following nuclear reaction

88Ra226  → 86Ra 222  Ra + 2He4

Q-value of
emitted α-particle = (Sum of initial mass − Sum of final mass) c 2 Where, c = Speed of
light It is given that:

m ( 88Ra226  )  =226.02540 u

m(86Rn222) =222.01750 u

m(2He4) =4.002603 u

Q-value =[ 226.02540 – (222.01750 + 4.002603) ] u C2

=0.005297 u C2

But  1 u =  931.5 MeV / C2

Q=  0.005297 × 931.5  ≈ 4.94 MeV

Kinetic energy of the α – particle = ( Mass number after decay /Mass number before decay)  × Q

=( 222/226 ) ×4.94 =4.85 MeV

b)Alpha particle decay of ( 86Rn 220 

86Rn 220   84 Po 216  + 2He4

It is given that:

Mass  ( 86Rn 220 )  of = 220.01137u

Mass  (84 Po 216 )   of =216.00189u

∴ Q – value = [  220.01137 – ( 216.00189+ 4.00260) ] ×931.5

≈ 641 MeV

Kinetic energy of the α-particle (  ( 220 – 4 ) / 220 )  × 6.41

=6.29MeV

13.The radionuclide 11C decays according to  6C11 5 B 11  + e  + v  : T1/2 = 20.3 min 

The maximum energy of the emitted positron is 0.960 MeV.

Given the mass values: m ( 6C11 ) = 11.011434 U and m ( 6B 11 ) = 11.009305 u

Calculate Q and compare it with the maximum energy of the positron emitted

The given nuclear reaction is:

6 C 115B11 + e+  + v

Half life of 6 C 11  nuclei, T1/2 =20.3 min

Atomic masses are: 

m( 6C 11 )  =11 .011434 u

m( 6B11 ) = 11.009305u

Maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q – value (Δ Q ) of the nuclear masses of the 6C 11

Δ Q = [ m’ (6C 11 ) – [ m’ (6B11 ) + me ] ] c2  …..(1)

Where,

me = Mass of an electron or positron = 0.000548 u

c = Speed of light

m’ = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add 6 me in the case of 11C and 5 me in the case of 11B.

Hence, equation (1) reduces to:

Δ Q = [ m (6C 11 )  – m(6B11 ) – 2m] c2

Here,m (6C 11 ) and m(6B11 )  are the atomic masses.

∴ Δ Q = [ 11.011434 – 11.009305 – 2 ×0.000548] c2

=(0.001033 c) u

But 1 u = 931.5 MeV / c2

ΔQ = 0.001033 × 931.5  ≈ 0.962 MeV

The value of Q is almost comparable to the maximum energy of the emitted positron

14.The nucleus 10 Ne23 decays by β emission. Write down the βdecay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

m ( 10 Ne23 )  =22.994466 u 

m(11 Na23 ) =22.989770 u

In β –  emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.β  emission of the nucleus 

                                                    _

10 Ne 23 10Ne2311Na23 + e + v +Q

It is given that:

Atomic mass m (10Ne23)  of = 22.994466 u

Atomic mass m (11Na23 ) of = 22.989770u

Mass of an electron , me = 0.000548 u

Q- value of the given reaction is given as:

Q = [ m(10Ne23) – [ m (11Na23 ) + me ] ] c2

There are 10 electrons in 10Ne23  and 11 electrons in 11Na23  . Hence, the mass of the electron is cancelled in the Q value  equation. 

∴ Q = [ 22.994466 – 22.9897770 ] c2

= (0.004696c 2) u

But  1 u= 931.5 MeV /c2

∴Q =0.004696 × 931.5  = 4.374 MeV

The daughter nucleus is too heavy as compared to e and v . Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e., 4.374 MeV

15.The Q value of a nuclear reaction A + b → C + d is defined by Q = [ mA+ mb− mC− md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.

(i) 1 H + 1 H3  → 1 + 1 

(ii) 6C12  + 6 C12 10Ne 20 + 2He4

Atomic masses are given to be

m( 1H2) = 2.014102 u 

m(1 H3  ) = 3.016049 u 

m(6 C12) = 12.000000 u

m( 10Ne 20) = 19.992439 u

(i) The given nuclear reaction is:

1H1 + 1H31H2 + 1H2

Atomic mass m(1H1) =1.0 07825 u

Atomic mass m (1H3 ) = 3.0164049 u

Atomic mass m (1H2 ) =  2.014102 u

According to the question, the Q-value of the reaction can be written as:

Q = [ m(1H1) + m (1H3 ) – 2m (1H2 ) ] c2

= [1.007825+3.016049-2×  2.014102] c2

Q= (-0.00433c2) u

But 1 u = 931.5 MeV /c2

∴Q = – 0.00433×931.5  = – 4.0334 MeV

The negative Q-value of the reaction shows that the reaction is endothermic.

(ii)The given nuclear reaction is : 6C12   +6C1210Ne20+ 2He4

It is given that:

Atomic mass of m(6C12) = 12.0 u

Atomic mass of m( 10Ne12 ) = 19.992439 u

Atomic mass of m(2He4) = 4.002603 u 

The Q-Value of this reaction is given as:

Q- [ 2m (6C12) -m(10Ne20) – m(2He4)] c2

= [ 2 ×12.0 – 19.992439 -4.002603 ]c2

= (0.004958  c2) u 

=0.004958 × 931.5 = 4.618377 MeV

The positive Q-Value of the reaction shows that the reaction is exothermic.

16.Suppose, we think of fission of a 26Fe56nucleus into two equal fragments, 13Al 28Is the fission energetically possible? Argue by working out Q of the process. Given m(26Fe56) = 55.93494 u and m(13Al 28) =27.9819 u

The fission of 26Fe56can be given as: 

26Fe56  13Al28

It is given that:

Atomic mass of m ( 26Fe56) = 55.93494 u

Atomic mass of m ( 13Al28) = 27.98191 u

The Q-value of this nuclear reaction is given as:

Q= [ m(26Fe56)  -2m ( 13Al28) ] c2

= [55.93494 -2 ×27.98191 ] c2

= (-0.02888 c2)u

but 1 u = 931.5 MeV /c2

∴Q = – 0.02888 × 931.5 = -26.902 MeV 

The Q-value of the fission is negative. Therefore, the fission is not possible energetically. For an energetically-possible fission reaction, the Q-value must be positive.

17.The fission properties of  94 Pu 239  are very similar to those of 92 U235 .The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure94 Pu 239  undergo fission?

Average energy released per fission of  94Pu239  , Eav =180 MeV

Amount of pure  94Pu239 , m =1 kg = 1000g

NA = Avogadro number = 6.023 × 1023

Mass number of  94Pu239  =239 g

1 mole of  94Pu239  contains NA atoms.

∴ gm of 94Pu239   contains ( ( NA / Mass number ) × m ) atoms

= (6.023 × 10 23 / 239  ) × 1000 = 2.52 × 10 24  atoms

∴ Total energy released during the fission of 1 kg of  94Pu239     is calculated as:

E = Eav  × 2.52 ×10  24

= 180 × 2.52 × 10 24  = 4.536 ×10 26 MeV

Hence, 4.536 ×10 26 MeVis released if all the atoms in 1 kg of pure 94Pu239 undergo fission

18.A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much 92 U235 did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of 92 U235 and that this nuclide is consumed only by the fission process.

Half life of the fuel of the fission reactor, t 1/2 =5years

=5 × 365 × 24×60×60 s

We know that in the fission of 1 g of  92U235  nucleus, the energy released is equal to 200MeV. 1 molw,i,e ., 235 g of 92U235   contains 6.023 × 10 23Atoms

∴ 1 g 92U235    6.023 × 10 23 / 235 atoms contains×

The total energy generated per gram of  92U235   is calculated as:

E=( 6.023 × 10 23 / 235 ) × 200MeV /g

=200 ×6.023 × 10 23 ×1.6 × 10 -19 ×106 /235  =  8.20 ×1010 J/g

The reactor operates only 80% of the time.

Hence, the amount of 92U235  consumed in 5 years by the 1000 MW fission reactor is calculated as:

(5 × 80 × 60×60×365×24×1000×10/ 100 ×8.20 × 1010 ) g

≈1538 kg

Initial amount of 92U235   = 2 × 1538 =3076kg

19.How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as

1H2 +1H2  → 2He3  ++n+ 3.27 MeV

The given fusion reaction is:

1H21H22He 3 +n+ 3.27 MeV

Amount of deuterium, m = 2kg

1 mole, i.e., 2 g of deuterium contains  6.023 ×1023  atoms

∴2.0 kg of deuterium contains  = (6.023 ×1023 /2  ) 2000 = 6.023 ×1026atoms

It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.

∴Total energy per nucleus released in the fusion reaction:

E = (3.27 /2  ) 6.023 ×1026 MeV

=( 3.27 /2 ) 6.023 ×1026  × 1.6 × 10 -19   ×10 6

=1.576 × 10 14 J

Power of the electric lamp ,p = 100W =100 J/s

Hence, the energy consumed by the lamp per second = 100 J

The total time for which the electric lamp will glow is calculated as:

=( 1.576 × 10 14/ 100 )s

1.576 × 10 14 / 100  × 60 ×60 × 24 ×365    = 4.9 × 10 4 years

20.Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0  fm.)

When two deuterons collide head-on, the distance between their centres, d is given as

Radius of 1st deuteron + Radius of 2nd deuteron

Radius of a deuteron nucleus = 2fm =2 ×10 -15 m

d =2 ×  10 -15 + 2 ×10-15 =4 × 10 -15 m

Charge on a deuteron nucleus = Charge on an electron =e =1.6 ×10 -19 C

Potential energy of the two-deuteron system

V = e 2/4 πε0 d

Where

ε=permittivity of free space

1/4 πε= 9 ×1o 9 Nm2C-2

∴V = 9 ×10  × (1.6 × 10 19)2 /4 ×1015 J

= 9 ×10 9 × (1.6 × 10 -19)2 /4 × 10 -15 ×(1.6 × 10 -19) eV

=360 MeV

Hence, the height of the potential barrier of the two-deuteron system is 360 keV.

21.From the relation R = R0 A1/3, where R0  is a constant and A is the mass number of a nucleus,show that the nuclear matter density is nearly constant (i.e. independent of A). 

We have the expression for nuclear radius as:

R =R0 A1/3

Where,
R0 = Constant.
A = Mass number of the nucleus

Nuclear matter density , ρ = Mass of the nucleus /Volumeof the nucleus

Let m be the average mass of the nucleus.

Henc

e, mass of the nucleus = mA

ρ = mA/(4/3  πR3)  = 3mA/4π(R0 A 1/3)3 = 3mA / 4 πR03 A = 3m / 4 πR03 

Hence, the nuclear matter density is independent of A. It is nearly constant

22.For the β+(positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K− shell, is captured by the nucleus and a neutrino is emitted)

e ++ AZ X → AZ-1 Y +v

Show that if β+ emission is energetically allowed, electron capture is necessarily allowed but  not vice−versa.

Let the amount of energy released during the electron capture process be Q1. The nuclear reaction can be written as:

e ++ AZ X → AZ-1 Y +v +Q1  …….(i)

Let the amount of energy released during the positron capture process be Q2. The nuclear

reaction can be written as:

e ++ AZ X → AZ-1 Y +v +Q2   ……..(ii)

mN (z XA) = Nuclear mass of  z XA

mN ( z-1  Y A )  = Nuclear mass of   z-1 Y A

m(z XA) = Atomic mass of  z XA

m(z-1 Y A) Atomic mass of z-1 Y

me = Mass of an electron
c = Speed of light

Q-value of the electron capture reaction is given as:

Q1 = [ mN (z XA) +me  –  mN (z-1 Y ) ] c2

= [m(z XA) – Zme   + me   – m (z-1 Y A) + (z-1)me  ] c2

= [ m(z XA) -m(z-1 Y A)] c2     ……….(iii)

Q-value of the positron capture reaction is given as:

Q=[mN(z XA) – mN (z-1 Y A) – me ] c2

= [m(z XA) – Zm -m (z-1 Y A) + (z-1) m-me  ] c2

= [m(z XA) – m(z-1 Y A) – 2me  ] c2  ……(iv)

It can be inferred that if Q2 > 0, then Q1 > 0; Also, if Q1 > 0, it does not necessarily mean that Q2 > 0.

In other words, this means that if β+ emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice versa. This is because the Q-value must be positive for an energetically-allowed nuclear reaction.

23.In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are  12 Mg 24 (23.98504u), 12 Mg 25 (24.98584u) and 12 Mg 26  (25.98259u)The natural abundance of  12 Mg 24  is 78.99% by mass. Calculate the abundances of other two isotopes.

Average atomic mass of magnesium, m =24.312 u

Mass of magnesium 12 Mg 24 isotope, m= 23.98504 u
Mass of magnesium 12 Mg 25 isotope, m2 = 24.98584 u

Mass of magnesium 12 Mg 26 isotope, m3 = 25.98259 u

Abundance of , 12 Mg 24  ,n  η =78.99%

Abundance of , 12 Mg 25  ,n2   η=x%

Hence, abundance of 12 Mg 26  , η3= 100 -x – 78.99% = (21.01 – x)%

We have the relation for the average atomic mass as:

m = m1η1 + m2η2  + m3η3  /  η12

24.312 = 23.98504  × 78.99 + 24.98584 × x + 25.98259 × (21.01 -x) /100

2431.2 = 1894.5783096 + 24.98584x + 545.8942159 – 25.98259 x

0.99675 x = 9.2725255

∴ x ≈ 9.3 %

and 21.01 -x = 11.71%

Hence, the abundance of 12 Mg 25  is  9.3% and that of
12 Mg 26 is 11.71%

24.The neutron separation energy is defined as the energy required to remove  a neutron from the nucleus. Obtain the neutron separation energies of the nuclei 20Ca41 and13 AI27from the following data:

m(20 Ca 40) = 39 .962591 u

m(20 Ca 41) = 40.962278 u 

m(13 AI26) = 25.96895u

For a neutron removal from 20 Ca 41   nucleus, the corresponding nuclear reaction can be written as:

20 Ca 41  20 Ca 40  + 0n

It is given that:

m(20 Ca 40  ) mass = 39.962591 u

m(20 Ca 41 ) mass =40.962278 u

m( 0n) mass =1.00866 u

The mass defect of this reaction is given as:

Δ m =m (20 Ca 40) + (0n) – m (20 Ca 41)

=39.962591 +1.008665 –  40.962278= 0.008978 u

but 1 u = 931.5 MeV /c2

∴Δm =0.008978 × 931.5 MeV /c2

Hence, the energy required for neutron removal is calculated as:

E =Δmc2

= 0.008978 × 931.5 =  8.363007 MeV

For 13Al 27  , the neutron removal reaction can be written as:

13Al 213Al 26 + 0n

It is given that:

m(13Al 27) mass= 26.981541 u

m(13Al 26 )mass= 25.986895 u

The mass defect of this reaction is given as:

Δm =m (13Al 26) + m(0n) – m(13Al 27)

=25.986895 +1.008665 -26.981541

=0.014019 u 

=0.014019 × 931.5 MeV /c2

Hence, the energy required for neutron removal is calculated as:

E =Δmc2

=0.014019 ×931.5 = 13.059MeV

25. A source contains two phosphorous radio nuclides 15P32 (T1/2 = 14.3d) and 15P33 (T1/2 = 25.3d) Initially, 10% of the decays come from 15P33  . How long one must wait until 90% do so?

Half life of 15P32 , T 1/2 = 14.3 days

Half life of 15P33 , p , T 1/2   = 25.3 days nucleus decay is10% of the total amount of decay

The source has initially 10% of 15P32   nucleus and 90% of 15P33   nucleus

Initially: 

Number of 15P33  nucleus = N 

Number of 15P32  nucleus = 9N

Finally:

Number of 15P33  nucleus = 9N’

Number of 15P32  nucleus = N’

For    15P32    nucleus, we can write the number ratio as:

N’/9N =(1/2) t/T1/2

N’ =9N(2) -t/14.3  ……(i)

For 15P33  ,we can write the number ratio as:

9N’ /N = (1/2) 1/T ‘ 1/2 

9N’ = N(2) -t/25.3   ……..(ii)

On dividing equation (1) by equation (2), we get:

1/9 = 9 ×2 ( t/25.3  –  t/14.3)

1/81 =2 (11 t/25.3  ×  14.3)

log 1 – log 81 = (-11 t/ 25.3 × 14.3 ) log 1

-11 t/25.3 × 14.3   = o- 1.908/0.301

t = 25.3 × 14.3× 1.908 / 11 ×0.301    

≈208.5 days

Hence, it will take about 208.5 days for 90% decay of  15P33

26. Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:

88Ra 223 →  82Pb 209    +   6 C14

88Ra 223 →  86Rn 219    +   2He4

Calculate the Q-values for these decays and determine that both are energetically allowed

Take a 6 C14  emission nuclear reaction: 

88 Ra 223  82 Pb 209 + 6 C14 

We know that:

Mass of  88 Ra 223  ,m1 = 223.01850 u

Mass of  82 Pb 209 , m2 = 208.98107 u 

Mass of  6 C14  , m3 = 14.00324 u 

Hence, the Q-value of the reaction is given as:

Q= (m1 -m2 -m3)c2

= (223.01850 – 208.98107  –  14.00324)c2

=(0.03419 c2 ) u

but 1 u = 931.5 MeV /c2

∴Q = 0.03419 931.5

= 31.848MeV

Hence, the Q-value of the nuclear reaction is 31.848 MeV. Since the value is positive, the reaction is energetically allowed.

Now take a 2He emission nuclear reaction:

88 Ra 223  86Rn219 +2He4

We know that

Mass of 88 Ra 223   ,m1 = 223.01850

Mass of 82Rn219  ,m2 =219.00948

Mass of 2He ,m3 = 4.00260

Q-value of this nuclear reaction is given as: 

Q =(m1 -m2 -m3)c2

=(223.01850 – 219.00948 – 4.00260 ) c2

=(0.00642  c2) u

=0.00642 ×  931.5 = 5.98MeV

Hence, the Q value of the second nuclear reaction is 5.98 MeV. Since the value is positive, the reaction is energetically allowed.

27.Consider the fission of 92 U 238 by fast neutrons. In one fission event, no neutrons are emitted and  the final end products, after the beta decay of the primary fragments, are 58 Ce 140  and  44Ru 99 Calculate Q for this fission process. The relevant atomic and particle masses are

m(92 U 238 ) = 238.05079 u

m(58 Ce 140 ) = 139.90543 u

m(44Ru 99  )=98.90594 u

In the fission of 92 U 238 10 β – particles decay from the parent nucleus. The nuclear reaction can be written as:

92 U 238 + 0n1 →  58Ce 140  44 Ru 99+  10 0  -1 e

It is given that:

Mass of a nucleus m1 , 92 U 238  = 238.05079 u

Mass of a nucleus m2 , 58Ce 140  = 139.90543 u

Mass of nucleus ,m3 44 Ru 99= 98.90594 u

Mass of a neutron m4 0n1 = 1.008665 u

Q-value of the above equation

Q= [ m’ ( 92 U 238) + m (0n1) – m’ (58Ce 140) -m’ (44 Ru 99) -10 me]c2

Where,
m’ = Represents the corresponding atomic masses of the nuclei

m’ ( 92 U 238) =m1 -92me

m’ ( 58Ce 140) =m2 -58me

m’ ( 44 Ru 99) =m3 -44me

m(0n1) =m4

Q= [ m1 -92m+ m4 -m2 +58me -m3 +44me -10me]c2

=[ m1+m4– m2-m3]c2

[238.0507 +1.008665 – 139.90543 – 98.90594]c2

=[0.247995 c2 ]u

but 1 u = 931 MeV/C2

∴Q=0.247995 ×931.5 =231.007MeV

Hence, the Q-value of the fission process is 231.007 MeV.

28.Consider the D−T reaction (deuterium−tritium fusion) 1 H2 + 1H32He 4 +n 

(a) Calculate the energy released in MeV in this reaction from the data

m( 1 H2) =2.014102 u

m(1H3) = 3.016049 u

b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what  temperature must the gas be heated to initiate the reaction? (Hint: Kineticenergy required for one fusion event = average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.)

Solution 28:
(a) Take the D-T nuclear reaction:  1H2 + 1H32He4 +n

It is given that:

Mass of 1H, m1 = 2.014102 u

Mass of 1H3 , m2 =3.016049  u

Mass of2He4 , m3 = 4.002603 u

Mass of0n1 , m3 = 1.008665 u

Q-value of the given D-T reaction is:

Q= [2.014102 + 3.016049 – 4.002603 – 1.008665] c2 = [0.018883  c2]u

but 1 u = 931.5 MeV /c2

∴Q= 0.018883 × 931.5 = 17.59 MeV

(b) Radius of deuterium and tritium, r ≈  2.0 fm = 2 ×10 -15 m

Distance between the two nuclei at the moment when they touch each other,

d= r+r = 4 × 10-15 m

Charge on the deuterium nucleus = e
Charge on the tritium nucleus = e
Hence, the repulsive potential energy between the two nuclei is given as:

V = e2 / 4πε0 (d)

Where,

ε= Permittivity of free space 

1/4πε = 9 ×10 Nm2 c-2 lim x→∞

∴ V =9 × 109 × (1.6 × 10 -19 ) / 4 ×10 -15  = 5.76 × 10 -14 J

= 5.76 × 10 -14 / 1.6 × 10 -19  = 3.6 × 10 5 eV =360keV

Hence, 5.76 × 10 -14 J  or r keV of kinetic energy (KE) is needed to overcome the coulomb repulsion between the two nuclei. 

However, it is given that

KE= 2 ×(3K T/2)

Where,
k =Boltzmann constant = 1.38 × 10 -23 m2 kg s-2 k-1

T= Temperature required for triggering the reaction

∴ T =KE/3K 

= 5.76 ×10 -14 / 3 × 1.38 × 10 -23 =1.39 ×10 K

Hence, the gas must be heated to a temperature of 1.39 ×10 Kto initiate the reaction.

29.Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ – decays in the decay scheme shown in figure. You are given that  

m(198 Au) =197.968233u

m(198 Hg) =197.966760 u

It can be observed from the given γ -decay diagram that γ 1 decays from the 1.088 MeVenergy level to the 0 MeV energy level

Hence, the energy corresponding to γ 1 -decay is given as:

E1 = 1.088 – 8 =1.088 MeV hv 1= 1.088 ×1.6 × 10-19  × 10 6

Where,
 h =Planck’sconstant = 6.6 × 10 -34 JSV= Frequency of radiation radiated by γ 1 decay

∴ v= E1/h

= 1.088 × 1.6 ×10 -19  × 10 / 6.6 × 10 -34 = 2.637 × 10 20 Hz

It can be observed from the given γ-decay diagram that γ2  decays from the 0.412 MeV energy level to the 0 MeV energy level.
Hence, the energy corresponding to  γ -decay is given as:

E= 0.412 – 0 = 0.412MeV hv2=0.412 ×1.6 ×10 -19  ×106 J

Where,

v= Frequency of radiation radiated by γ2 – decay

∴ v2 = E2/h

=0.412 × 1.6 × 10 -19  × 10/ 6.6 × 10 -34  = 9.988 ×10 19 Hz

It can be observed from the given γ -decay diagram that γ3 decays from the 1.088 MeV energy level to the 0.412 MeV energy level.
Hence, the energy corresponding to γ3 -decay is given as:

E3 = 1.088 -0.412 =0.676 MeV hv = 0.676 × 10 -19  ×10 6

where,

v= Frequency of radiation radiated by γ3  decay

∴ v= E3/h 

= 0.676×1.6 ×10 -19  × 10 /6.6 × 10 -34   = 1.639 × 10 20 Hz

Mass of m (78 Au 198) = 197 .968233 u

Mass of m( 80 Hg 198) = 197.966760 u

 1 u = 931 .5 MeV /c2

Energy of the highest level is given as:

E= [m((78 Au 198) – m ( 80 Hg 198) ]

= 197.968233  –  197.966760 = 0.001473 u

= 0.001473 × 931.5 = 1.3720995MeV

β1 decays from the 1.3720995 MeV level to the 1.088 MeV level

Maximum kinetic energy of the β  particle = 1.3720995 − 1.088

= 0.2840995 MeV

β decays from the 1.3720995 MeV level to the 0.412 MeV level

Maximum kinetic energy of the β 2 particle = 1.3720995 – 0.412

=  0.9600995 MeV

30.Calculate and compare the energy released by
(a) fusion of 1.0 kg of hydrogen deep within Sun and

(b) the fission of 1.0  -kg 235 U  in a fission reactor.

(a) Amount of hydrogen, m = 1 kg = 1000 g

1 mole, i.e., 1 g of hydrogen (1H1) contains 6.023  × 1023 Atoms

∴ 1000 g of 1H1 contains 6.023 × 10 23  × 1000 atoms 

Within the sun, four 1H1   nuclei combine and form one   2He4 nucleus. In this process 26MeV of energy is released.

Hence, the energy released from the fusion of 1 kg 1H1    is 

E=6.023  ×1023  ×26 × 10 /4

=39.1495 × 10 26 MeV

b)Amount of 92 U 235  =1000gm 

1 mole, i. e., ,235 g of 92 U 235  contains  6.023 × 10 23 atoms

∴ 1000g of 92 U 235   contains 6.023  ×1023  × 1000 /235  atoms

It is known that the amount of energy released in the fission of one atom of  92 U 235   200 MeV

Hence, energy released from the fission of 1kg of 92 U 235   is 

E = 6 × 10 23 × 1000 ×200 /235 

= 5.106 ×1026  MeV 

E1/E2 = 39.1495 × 10 26/ 5.106 × 10 26 =7.67 ≈ 8

Therefore, the energy released in the fusion of 1 kg of hydrogen is nearly 8 times the energy released in the fission of 1 kg of uranium

31. Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of 235U to be about 200MeV.

Amount of electric power to be generated, p =2  × 10  MW 10% of this amount has to be obtained from nuclear power plants.

Amount of nuclear power, p=( 10 /100 ) × 2 × 10 5

= 2× 10 MW

=2 × 10 ×10  J/s

= 2 ×10 10 × 60 ×60 ×24 ×365 J/y

Heat energy released per fission of a  235 U nucleus,  E =200 MeV

Efficiency of a reactor =25%

Hence, the amount of energy converted into the electrical energy per fission is calculated as:

(25/100 ) 200 = 50 MeV

=50 × 1.6 ×10 -19  × 10 =8 ×10 -12 J

Number of atoms required for fission per year:

2 × 1010 ×60 ×60×24 ×365 / 8 ×10 -12 = 78840 × 10 24  atoms

1 mole, i.e., 235g of U 235 contains  6.023 × 10 23  Atoms

∴Mass of 6.023 × 10 23 atoms of U 235 =235g = 235  × 10 -3 kg

∴ Mass of  78840 × 10 24 atoms of U 235

= (235 × 10 -3 / 6.023 ×10 23 ) 78840  × 10 24

= 3.076 × 10 kg 

Hence, the mass of uranium needed per year is 3.076 × 10 kg 

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